13.10.2012 Views

boylistad

boylistad

boylistad

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

918 ⏐⏐⏐ RESONANCE<br />

Therefore,<br />

and<br />

Qp � � �20<br />

XL � 2pfpL � 2p(50 kHz)(1 mH) � 314 �<br />

Ql � � �31.4<br />

Rp � Q 2 lR � (31.4) 2 (10 �) � 9859.6 �<br />

Qp � � �20 (from above)<br />

so that � 6280<br />

resulting in Rs � 17.298 k�<br />

However, the source resistance was given as 40 k�. We must therefore<br />

add a parallel resistor (R′) that will reduce the 40 k� to approximately<br />

17.298 k�; that is,<br />

� 17.298 k�<br />

Solving for R′:<br />

R′ � 30.481 k�<br />

The closest commercial value is 30 k�. At resonance, XL � XC, and<br />

XC �<br />

C � �<br />

and C � 0.01 mF (commercially available)<br />

ZTp � Rs � Q 2 fp 50,000 Hz<br />

� ��<br />

BW 2500 Hz<br />

XL 314 �<br />

� �<br />

Rl 10 �<br />

R Rs � 9859.6 �<br />

� ��<br />

XL 314 �<br />

(Rs)(9859.6) ��<br />

Rs � 9859.6<br />

(40 k�)(R′)<br />

��<br />

40 k��R′<br />

1<br />

�<br />

2pfpC<br />

1<br />

1<br />

� ��<br />

2pfp XC 2p(50 kHz)(314 �)<br />

lRl � 17.298 k� � 9859.6 �<br />

� 6.28 k�<br />

with V p � IZ Tp<br />

Vp 10 V<br />

and I �����1.6 mA<br />

ZTp 6.28 k�<br />

The network appears in Fig. 20.39.<br />

I 1.6 mA<br />

R s 40 k� R� 30 k�<br />

R l<br />

10 �<br />

L 1 mH<br />

FIG. 20.39<br />

Network designed to meet the criteria of Fig. 20.38.<br />

ƒ r<br />

C 0.01 mF

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!